Question #128093
A 500-ton train and collides with a buffer stop while traveling at 2 m/s. A buffer stop is a safety device and is designed to stop trains in case of brake failures. The mass of the buffer stop is 1.5 tons. It acts as a spring with a spring constant k = 1 x 106 N/m.

a. Estimate the mean deceleration, in [m/s2], experienced by the train as it is halted.

b. Estimate the distance, in [m], by which the buffer stop spring is compressed.
1
Expert's answer
2020-08-03T15:28:49-0400

The kinetic energy of the train got stored as elastic potential energy of the spring as a result of stopping it. So, if the spring gets compressed by a distance of x and it's force constant is k, its elastic potential energy stored will be 0.5kx20.5kx^{2}. And if the train of mass m was moving with a speed of v, then its kinetic energy was 0.5mv20.5mv^{2} . So, equating both, we get:

0.5mv2=0.5kx20.5mv^{2} = 0.5kx^{2}

0.5×500×103×22=0.5×1×106×x20.5\times500\times10^{3}\times2^{2}=0.5\times1\times10^{6}\times x^{2}

x=1.41m

Answer b) The distance, in [m], by which the buffer stop spring is compressed = 1.41m

Then we can calculate the deceleration of the train:

v2=u2+2asv^{2}=u^{2}+2as , where v - final speed, u - initial speed, a - acceleration, s - distance.

02=22+2×a×1.410^{2}=2^{2}+2\times a\times 1.41

a=-1.41ms21.41\frac{m}{s^{2}}

Answer a) Deceleration 1.41ms21.41\frac{m}{s^{2}}


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